UA-CHEM 127: Advanced General Chemistry I Physical principles underlying chemistry

I. EXPERIMENT AND THEORY IN MODERN CHEMISTRY

Research in chemistry proceeds along two directions: Experimental and theoretical. Experimental chemistry is the science of observation. We pose a question to nature, carry out the corresponding experiment, and observe the outcome. What happens if we mix two substances? If we blast a molecule with a laser, what will it do? If we introduce a biomimetic into a living organism, how will it respond? Experiments produce data that relate to each individual observation. Theory, on the other hand, is concerned with building general frameworks for placing large numbers of individual observations into a rational order. It does so by building mathematical models, deducing general laws, and using these to predict the outcomes of future experiments. The tools of theory are usually mathematics and physics applied to chemical problems. Without theory, experimental science is just a vast catalog of unrationalized observations. Without experiment, theory is simply a long list of conjectures with no verification or falsification. Keeping this juxtaposition in mind, we briefly review the basic laws of chemical transformations, keeping in mind their historical origins and the observations out of which they grew. II. FUNDAMENTAL LAWS OF CHEMICAL REACTIONS AND CHEMICAL EQUATIONS

1. The law of conservation of mass (Lavoisier, 18th century): Lavoisier was one of the first to carry out quantitatively accurate chemical measurements. He demonstrated that combustion required oxygen, and he demonstrated oxygen’s role in the rusting of metals. His observations led him to deduce the following general law known as the law of conservation of mass: In every chemical transformation, an equal quantity of matter exists before and after the reaction.

(Because he was a tax collector and nobleman, Lavoisier was branded a traitor during the French Revolution and beheaded in 1794.) 2. law of definite proportions (Joseph Proust, shortly after Lavoisier): Proust studied metal compounds, including metal oxides, carbonates and sulfides. From the work of Robert Boyle in the 17th century, it was understood that substances that could be broken down into more fundamental components were mixtures or compounds. Substances that could not be further broken down were referred to as elements. Thus, Proust deduced the so-called law of definite proportions: In a given chemical compound, the proportion by mass of the elements that compose it are fixed, independent of the origin of the compound or its mode of preparation. This is basically saying that sodium chloride, for example, is always NaCl, no matter how it is obtained, made, or prepared. There are no “intermediate” compounds. 3. The law of multiple proportions: (John Dalton, shortly after Proust): Studied gases and gaseous mixtures under different external conditions. Building on Proust’s work, he noted that mathematically discrete manner in which elements combined to form different compounds. For example, in carbon monoxide (CO), the mass ratio of oxygen to carbon mO /mC = 1.33 and in carbon dioxide (CO2 ) mO /mC = 2.66. Thus, since the amount of carbon is fixed in each compound, we can look at how the amount of oxygen varies, and we find that mO (in CO2 )/mO (in CO) = 2. The generalization of this idea is the law of multiple proportions: When two elements form a series of compounds, the masses of one that combine with a fixed mass of the other are in the ratio of (small) integers to each other.

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This law is obeyed by all gaseous compounds, which is what Dalton studied. Certain solids are exceptions to both this rule and the law of definite proportions. An example is the solid w¨ ustite, which can range from Fe0.95 O to Fe0.85 O, depending on the mode of preparation. These formulae express the incommensurate compositions possible in this solid. But, since atoms are essentially indestructible (we have to work hard to get them to fission!), this explains why we do not find compounds such as C13/7 H5/3 in nature. Dalton’s observations led him to propose the notion of the atom as the fundamental and indestructible building blocks of matter. 4. The law of combining volumes (Joseph Gay-Lussac, Amedeo Avagadro, Stanislao Cannizzaro, early 19th century following Dalton): Again, based on studies of gases and how they combine, the law of combining volumes has been attribued to these three. The generalization of the observations is stated as follows: When two gases are allowed to react, such that the gases are at the same temperature and pressure, the volumes of each gas consumed will be in the ratio of small integers. Moreover, the ratio of the volume of each product gas to the volume of either reacting gas will be a ratio of simple integers. Example: 3 volumes of hydrogen + 1 volume of nitrogen → 2 volumes of ammonia. Although this is a statement about gases, its implications can be deduced: the coefficients in chemical equations expressions how much of different reactants combine to give products will be integers. 5. Avogadro’s hypothesis (1811): Equal volumes of different gases (at the same temperature and pressure) contain equal numbers of particles. Example: How many particles are in 1 L of O2 gas, and how many particles are in 1 L of H2 gas? The density of O2 is 1.429 g/L, and the density of H2 gas is 0.0899 g/L. Solution: mass of O2 = (1L)(1.429g/L) = 1.429g moles of O2 =

1.429g = 0.04466mol 31.998g/mol

moleculesofO2 = (0.04466mol)(6.022 × 1023 molecules/mol) = 2.69 × 1022 molecules

mass of H2 = (1L)(0.0899g/L) = 0.0899g moles of H2 =

0.0899g = 0.04466mol 2.0158g/mol

moleculesofH2 = (0.04466mol)(6.022 × 1023 molecules/mol) = 2.69 × 1022 molecules

Combining Avogadro’s hypothesis with the law of combining volumes allows the statement written above about volumes of hydrogen and nitrogen combining to give ammonia as a statement about definite numbers of molecules rather then volumes. This number could be expressed as moles, or even as individual molecules. The statement then reads: 3H2 + N2 −→ 2NH3 This is an example of a balanced chemical equation. What does it mean to be balanced? It means that the law of conservation of mass is obeyed – equal quantities of hydrogen and nitrogen exist on both sides of the reaction. Furthermore, the coefficients, i.e., the numbers multiplying H2 , N2 and NH3 in the above equation, are expressible as simple integers. When balancing chemical equations, this serves as an important check. 2

III. METHODS OF BALANCING CHEMICAL EQUATIONS

1. Balancing by inspection: Consider the example: NH4 NO3 −→ N2 O + H2 O We notice that there are 2 nitrogens on both the left and right sides, so nitrogen is already balanced. There are 3 oxygens on the left but only 2 on the right, so oxygen is not balanced. Similarly, there are 4 hydrogens on the left and only 2 on the right, so hydrogen is not balanced. The balancing of oxygen and hydrogen can be handled together by making the coefficient of H2 O equal to 2. Then, the balanced equation reads NH4 NO3 −→ N2 O + 2H2 O Now there are equal quantities of all elements on both sides of the reaction, and mass conservation is satisfied. Balancing by inspection in this way is quick and useful when you can see the solution easily. 2. Algebraic method: Consider the reaction: C4 H10 + O2 −→ CO2 + H2 O This equation could be balanced by inspection. However, we will use it to illustrate another approach – the algebraic approach. To balance a reaction algebraically, we start by putting unknown coefficients in front of each molecular species in the equation: xC4 H10 + yO2 −→ zCO2 + wH2 O Then we write down the balance conditions for each element in terms of the unknowns. In this case, there are are four unknowns, x, y, z and w. From carbon balance, we have the condition: 4x = z The hydrogen balance condition is: 10x = 2w and the oxygen balance condition gives: 2y = 2z + w This constitutes a set of only three equations for the four unknowns. The fourth condition comes from recognizing that chemical equations specify relative amounts of reactants and products. Thus, one of the coefficients is arbitrary, and all the others can be expressed relative to it. Thus, we may take one of the coefficients to be 1, arbitrarily, which then leaves only three unknowns and three conditions to determine them. The choice of which coefficient to set to 1 is completely arbitrary and may be taken to be the one that is most convenient for solving the problem. In this case, the choice x = 1 allows us to arrive at a solution almost immediately, for then it follows that z = 4 and w = 5. From the oxygen balance condition, y can be determined to be 2y = 8 + 5 = 13 so y = 13/2. Thus, the balanced equation could be written as C4 H10 +

13 O2 −→ 4CO2 + 5H2 O 2

This is a perfectly fine way of writing the equation, however, customarily, it is preferable to express all the coefficients as integers, in accordance with the law of combining volumes and Avogadro’s hypothesis. Thus, we 3

can multiply through the equation by 2 on both sides (it can be treated as an algebraic equation in this way) and obtain the final result: 2C4 H10 + 13O2 −→ 8CO2 + 10H2 O As a second example, consider the reaction: NaCl + SO2 + H2 O + O2 −→ Na2 SO4 + HCl Again, we put unknown coefficients in front of each molecular species: xNaCl + ySO2 + zH2 O + wO2 −→ uNa2 SO4 + vHCl Writing down the balance conditions on each element gives: Sodium balance: x = 2u Chlorine balance: x=v Sulfur balance: y=u Oxygen balance: 2y + z + 2w = 4u Hydrogen balance: 2z = v Setting u = 1 arbitrarily, gives the immediate solution: x=2

v=2

y=1

z=1

and 2 + 1 + 2w = 4 or w = 1/2. In order to clear the fraction, we multiply all the coefficients by 2 and write down the balanced equation: 4NaCl + 2SO2 + 2H2 O + O2 −→ 2Na2 SO4 + 4HCl

IV. LIMITING REAGENT

Chemical equations specify precise amounts of reactants that will combine to give products. In a real chemical reaction, arbitrary amounts of reactants are mixed together and allowed to form products, which means generally that one of the reactants will be used up completely, while excess amounts of other reactants will remain after the reaction has run to completion. The species that is fully consumed is called the limiting reagent. The limiting reagent will then determine what the yield of product will be. When the quantities of reactants that are allowed to combine are specified, identifying the limiting reagent amounts to determining which of the reacting species yields the smallest amount of product. Consider the following example. Sulfuric acid is produced in the reaction: 4

2SO2 + O2 + 2H2 O −→ 2H2 SO4 Suppose 400 g of SO2 , 175 g of O2 and 125 g of H2 O are mixed and allows to react. Which reactant is the limiting reagent? Solution: First convert the given masses to moles: moles of SO2 =

400 g = 6.24 mol SO2 64.06 g/mol

moles of O2 =

175 g = 5.47 mol O2 31.996 g/mol

moles of H2 O =

125 g = 6.94 mol H2 O 18.02 g/mol

From the reaction equation, we see that 2 moles of SO2 yields 2 moles of H2 SO4 or 1 mole of SO2 yields 1 mole of H2 SO4 or 6.24 moles of SO2 yields 6.24 moles of H2 SO4 Similarly, 1 mole of O2 yields 2 moles of H2 SO4 or 5.47 mole of O2 yields 10.94 moles of H2 SO4 and 2 moles of H2 O yields 2 moles of H2 SO4 or 1 mole of H2 O yields 1 mole of H2 SO4 or 6.94 moles of H2 O yields 6.94 moles of H2 SO4 Since SO2 yields the smallest quantity of product, it is the limiting reagent. V. PERCENTAGE YIELD

A balanced chemical equation tells what the theoretical or ideal yield of the reaction should be, assuming perfect completion of the reaction. In reality, the actual yield may be less than the theoretical yield. Reasons for this include: 1. There may be competing reactions which hinder the process under consideration. 2. External conditions may not be perfectly maintained. 3. Reactants may not be pure.

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The ratio of the actual yield to the theoretical yield is a quantity known as the fractional yield: fractional yield =

actual yield theoretical yield

and the percentage yield is obtained from the fractional yield according to percentage yield = 100% × (fractional yield) Example: Pure zinc can be produced in the following two-step process: 2ZnS + 3O2 −→ 2ZnO + 2SO2 ZnO + CO −→ Zn + CO2 Suppose 5.32 kg of ZnS yield 3.30 kg of pure zinc. What is the percentage yield? Solution: Convert the mass of ZnS to moles: moles ZnS =

5320 g = 54.6 moles ZnS 97.46 g/mol

How do we figure out how many moles of Zn are produced in the two-step reaction? We need to obtain the overall reaction, which is just a “sum” of the individual steps. We can obtain the equation of the overall reaction by adding the equations for the individual steps. However, this addition must be carried out in a particular way. Specifically, note that the species ZnO is produced in the first step and consumed in the second step. Thus, it does not appear among the final products of the process, and for this reason, it is called an intermediate. The equations for the individual steps must be added in such a way that all intermediates cancel out. Since adding reaction equations works like the adding of algebraic equations, this means that the coefficients of intermediates must be the same in the individual steps. In the above process the coefficients of ZnO are different in each step. But, if the first step is divided by 2, so that the reaction equation reads: 3 ZnS + O2 −→ ZnO + SO2 2 then ZnO has the same coefficient in the first step as is does in the second and the equations can now be added: 3 ZnS + O2 −→ ZnO + SO2 2 ZnO + CO −→ Zn + CO2 −−−−−−−−− −−−−−−−−− 3 ZnS + O2 + ZnO + CO −→ ZnO + SO2 + Zn + CO2 2 3 ZnS + O2 + CO −→ SO2 + Zn + CO2 2 where ZnO has been “cancelled” algebraically on both sides. From the overall reaction, we see that 1 mole of ZnS yields 1 mole of Zn or 54.6 moles of ZnS yields 54.6 moles of Zn Thus, the mass of Zn produced is mass of Zn = (54.6 mol) × (65.39 g/mol) = 3570 g = 3.57 kg This is the theoretical yield predicted by the reaction equations. The actual yield is only 3.30 kg, so that the fractional yield is fractional yield =

3.3 kg = 0.924 = 92.4% 3.57 kg 6

In multi-step processes, it is important that each step have as high a percentage yield as possible. The reason for this is that the overall percentage yield will be a product of the percentage yields of each step. As an example, suppose there is a 10-step process with a 50% yield for each step: Step 1 Step 2 ··· Step 10

percentage yield = 50% percentage yield = 50% percentage yield = 50%

The overall percentage yield would be the product of the individual percentages, or (0.5)10 = 0.001 = 0.1% Even if the percentage yield of each step were 90%, the overall yield would be (0.9)10 = 35%, which is better, but still somewhat low. If the percentage yield of each step were 95%, then the overall yield would be (0.95)10 = 60%, which might make the process feasible. VI. LAW OF CONSERVATION OF ENERGY

The law of conservation of energy is one of the basic laws of physics and therefore governs the microscopic motion of individual atoms in a chemical reaction. The law of conservation energy states: In a closed system, i.e., a system that isolated from its surroundings, the total energy of the system is conserved. In SI units, energy has units of Joules. 1 Joule = 1 kg·m2 ·s−2 . Some forms of energy: 1. Kinetic energy – energy of motion. 2. Potential energy – energy of “location” with respect to some reference point. 3. Chemical energy – energy stored in chemical bonds, which can be released in reactions. 4. Electrical energy – energy created by separating charges; energy stored in a battery, for example. 5. Thermal energy – energy given off as heat, such as friction. Since everything has a microscopic origin, the last three are really special cases of potential and kinetic energies, however, the classification is useful. The kinetic energy of an object of mass m, moving with a velocity v is given by KE =

1 m|v|2 2

Recall that |v| is the speed v of the particle, so that the kinetic energy can be written as 1/2mv 2 . Potential energy is a little less straightforward. Since it is an energy of location with respect to some reference point, the potential energy of an object depends on the specific situation. An example is gravity. An object of mass m at a height h above the Earth’s surface has a gravitational potential energy P E = mgh where g is the acceleration due to the Earth’s gravity. In this example, the potential energy has a simple form, namely that it depends linearly on the height h, which describes the object’s location and may even be varying in time if, for example, the object is actually falling toward the Earth. 7

In general, the potential energy may not be such a simple function of location. In this case, one needs a potential energy curve to describe the potential energy as a function of some coordinate describing the object’s location. Consider the example of a mass attached to a spring moving in one spatial dimension. Let x represent the mass’s position along the x-axis, and let x0 represent its equilibrium position. As the mass stretches the spring (x > x0 ), its potential energy increases, and as it compresses the spring compresses (x < x0 ), its potential energy increases as well. The potential energy can be described by a potential energy function V (x) that is symmetric about x = x0 , as represented in the figure below:

V(x) Tangent

x x1 FIG. 1.

Notice, also, that the mass actually moves under the action of a force, which also changes as a function of x. In fact, the force exerted by the spring on the mass can be determined from the potential energy curve via Force at x = −slope of the potential energy curve at x That is, the force is the slope of the line tangent to the curve at the point x, as shown in the figure above for the point x1 . In terms of a derivative, the force is given by F =−

dV dx

Note that, at the bottom of the gully, where the curve is flat, the force is 0. The SI units of force are Newtons. 1 N = 1 kg·m·s−2 . Consider the case of a diatomic molecule:

B

A

r FIG. 2.

The distance between the atoms A and B is r. The chemical forces that hold the molecule together come from a potential energy curve V (r) that depends only on the distance between them. Such a curve might look like:

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V(r)

r FIG. 3.

The presence of a deep “well” indicates a particular distance of lowest potential energy. This corresponds to the bond length of the molecule and is, therefore, the most likely value of the separation of the two atoms (think of a ball placed in a gully of this shape – if placed at the bottom of the well, the force on it would be 0, hence it would not move, but would remain there forever unless disturbed). According to Newton’s second law of motion, force is the action that causes a change in the motion of a particle or sets it in motion from rest. Consider a particle of mass m moving in one dimension so that its position is x. If a force F acts on the particle, it will be set in motion, so that x is no longer a fixed number but a function x(t) of time t. At any instant in time, the particle will have a velocity v defined to be the rate of change of x with respect to time v(t) =

dx dt

In general v is not fixed but is also a function v(t) of time t. If v(t) is not constant, in fact, then the particle also has an acceleration a defined as the rate of change of v with respect to time a(t) =

dv dt

a(t) =

d2 x dt2

Since v = dx/dt, it follows that

Newton’s second law of motion states that the force F acting on the particle is directly proportion to this acceleration, with the mass m as the constant of proportionality F = ma = m

dv d2 x =m 2 dt dt

Since motion generally occurs in three rather than one dimension, position, velocity, acceleration, and force are all vector quantities. The position is usually denoted r, and r has three components r = (x, y, z). Similarly velocity v has components v = (vx , vy , vz ), accleration a = (ax , ay , az ), and F = (Fx , Fy , Fz ). In vector notation, Newton’s second law reads F = ma = m

dv d2 r =m 2 dt dt

Hence, if r is not a fixed vector but a function r(t) of time, this means that each component of r is a function of time r(t) = (x(t), y(t), z(t)). Work: Consider again a particle of mass m moving in one spatial dimension. If a force F is needed to move the particle from position x1 to position x2 , then mechanical work W has been done on the particle. Since, as we have 9

seen, F can be a function F (x) of x, the general definition of work is the area under the force function F (x) between x1 and x2 , i.e. the integral Z x2 F (x)dx W = x1

However, since F (x) = −dV /dx, this becomes Z x2 dV dx = − [V (x2 ) − V (x1 )] = V (x1 ) − V (x2 ) W =− x1 dx Note that if work is done against a force F (x) in moving the particle from x1 to x2 , then that work, which we denote as W ′ , is −W or Z x2 F (x)dx = V (x2 ) − V (x1 ) W′ = − x1

Now if the velocity of the particle at x1 is v1 and at x2 , it is v2 , then by energy conservation 1 1 mv12 + V (x1 ) = mv22 + V (x2 ) 2 2 Rearranging this gives V (x1 ) − V (x2 ) =

1 1 mv 2 − mv 2 2 2 2 1

Hence, the work W is also equal to the change in kinetic energy W =

1 1 mv 2 − mv 2 2 2 2 1

a result known as the work-kinetic energy theorem.

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